HDU 3001 Travelling(状态压缩dp)

Travelling

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5765    Accepted Submission(s): 1858

Problem Description

After coding so many days,Mr Acmer wants to have a good rest.So travelling is the best choice!He has decided to visit n cities(he insists on seeing all the cities!And he does not mind which city being his start station because superman can bring him to any city at first but only once.), and of course there are m roads here,following a fee as usual.But Mr Acmer gets bored so easily that he doesn't want to visit a city more than twice!And he is so mean that he wants to minimize the total fee!He is lazy you see.So he turns to you for help.

 

Input

There are several test cases,the first line is two intergers n(1<=n<=10) and m,which means he needs to visit n cities and there are m roads he can choose,then m lines follow,each line will include three intergers a,b and c(1<=a,b<=n),means there is a road between a and b and the cost is of course c.Input to the End Of File.

 

Output

Output the minimum fee that he should pay,or -1 if he can't find such a route.

 

Sample Input

   
   
   
   

    
    
    
    2 1
1 2 100
3 2
1 2 40
2 3 50
3 3
1 2 3
1 3 4
2 3 10
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    100
90
7 
   
   
   
   

 

Source

2009 Multi-University Training Contest 11 - Host by HRBEU

 

题意:要走完所有的城市，每条边有一个代价，求走完所有的城市的最小代价，起初没看到每个城市最多可走两边。

思路:用二进制的状态压缩果然wa了，用三进制的状态压缩，先把进制表打好，然后用一个二维的数组表示出这个数每位在三进制下的状态，状态转移。

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<map>
#include<cstring>
#include<map>
#include<set>
#include<queue>
#include<stack>
using namespace std;
const double eps = 1e-10;
const int inf = 0x3f3f3f3f, N = 6e4;
typedef long long ll;
int dp[N][12], Map[12][12], n, m, three[12], vis[N][12];
void init()
{
    three[0] = 1;
    for(int i = 1; i<=10; i++)
        three[i] = three[i-1] * 3;
    for(int i = 0; i<three[10]; i++)
    {
        int temp = i;
        for(int j = 1; j<=10; j++)
        {
            vis[i][j] = temp % 3;
            temp /= 3;
        }
    }
}
int main()
{
    init();
    while(~scanf("%d%d", &n, &m))
    {
        memset(Map, 0x3f, sizeof(Map));
        int u, v, w;
        while(m--)
        {
            scanf("%d%d%d", &u, &v, &w);
            Map[u][v] = Map[v][u] = min(w, Map[u][v]);
        }
        memset(dp, 0x3f, sizeof(dp));
        int num = three[n] - 1;
        for(int i = 1; i<=n; i++)
            dp[three[i-1]][i] = 0;
        int ans = inf;
        for(int i = 1; i<=num; i++)
        {
            for(int j = 1; j<=n; j++)
            {
                if(dp[i][j] == inf)
                    continue;
                for(int k = 1; k<=n; k++)
                {
                    if(vis[i][k]>=2 || Map[j][k] == inf || j == k)
                        continue;
                    dp[i + three[k-1]][k] = min(dp[i + three[k-1]][k], dp[i][j] + Map[j][k]);

                }
            }
        }
        for(int i = 1; i<=num; i++)
        {
            int ok = 1;
            for(int j = 1; j<=n; j++)
                if(!vis[i][j])
                {
                    ok  = 0;
                    break;
                }
            if(ok)
                for(int j = 1; j<=n; j++)
                    ans = min(dp[i][j], ans);
        }
        if(ans != inf)
            cout<<ans<<endl;
        else cout<<-1<<endl;
    }
    return 0;
}